Answer to Question #147285 in Statistics and Probability for Ali Ahmed

Question

A car wholesaler estimates the quantity of new cars retailed that have been returned back a many
number of times for the rectification of defects during the warranty. The results are given in the
following table. Show complete working.
Number of cars
returns back
0 1 2 3 4
Probability 0.25 0.34 0.22 0.07 0.03
a. Compute the mean of the number of cars returns for rectifications for defects during the
warranty time.
b. Calculate the variance of the number of cars returns for rectifications for defects during
the warranty time.

Answer 1

a. Mean

$Mean=\sum_i XiPXi$

$=0(0.25)+1(0.34)+2(0.22)+3(0.07)+4(0.03)$

$=1.11$

b. Variance

. $Variance=\sum_i Xi^2PXi-\bar{X}^2$

$=0(0.25)+1(0.34)+4(0.22)+9(0.07)+16(0.03)-1.11^2$

$=2.33-1.11^2=1.0979$